Meromorphic functions on a Riemann surface as quotient of polynomials

algebraic-geometryriemann-surfaces

Consider a compact Riemann surface $C$ embedded in a projective plane with coordinates $[x:y:z]$.
Is it possible to write any meromorphic function $f\in M(C)$ as a fraction $f=\frac{P(x,y,z)}{Q(x,y,z)}$ with $P,Q$ homogeneous polynomials of the same degree having no common zero on $C$?
Of course $Q(x,y,z)$ will have zeroes on $C$ since a (non constant) meromorphic function on $C$ can not be holomorphic everywhere, but I'm asking whether at such zeroes $P$ will be non-zero.
I am pretty sure that Riemann-Roch guarantees that the answer is yes provided the common degree of $P$ and $Q$ is large enough, but I'd be grateful for a clean write-up.

Best Answer

Often no, by Bezout's theorem.

Let $$E:zy^2=x^3+z^3\subset \Bbb{P^2(C)}$$

  • For an homogeneous polynomial $f\in \Bbb{C}[x,y,z]$ of degree $d$ its zeros on $E$ are well defined, to define their multiplicities: take $w\in \{x,y,z\}$ not vanishing at $P$ and look at the order at $P$ of the zero of the meromorphic function $f/w^d$.

  • $x$ has 3 simple zeros at $[0:1:1],[0:-1:1],[0:1:0]$

  • $x^d$ has $3d$ zeros

  • $f/x^d$ is a meromorphic function, with the same number of zeros and poles, so $f$ must have $3d$ zeros.

  • The meromorphic function $x/z$ has two simple zeros at $[0:1:1],[0:-1:1]$ and a double pole at $[0:1:0]$.

It can't be that $$x/z=f/g$$ with $f,g$ homogeneous of same degree $d$ and with no common zero, as $f/g$ would have $3d$ zeros and poles whereas $x/z$ has $2$ zeros and poles.

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